4.1 An electrician, rushing to an urban area with a power outage, drives a truck of mass 1 350 kg towards the east travelling at 120 km.h-1 - NSC Technical Sciences - Question 4 - 2020 - Paper 1

To convert the truck's speed from km.h-1 to m.s-1, we use the conversion factor:

The initial momentum of the car can be calculated using the formula:

$p = m \cdot v$

Given that the mass of the car is 1 050 kg and its velocity is 16,67 m.s-1, the momentum is:

$p_{car} = 1 050 \text{ kg} \cdot 16,67 \text{ m.s}^{-1} = 17 503,5 \text{ kg.m.s}^{-1}.$

Thus, the initial momentum of the car before the collision is 17 503,5 kg.m.s-1.

The principle of conservation of linear momentum states that the total linear momentum of an isolated system remains constant if no external forces act upon it.

We must compare the total kinetic energy before and after the collision:

Initial kinetic energy (KE_initial):

$KE_{initial} = \frac{1}{2} m_{truck} v_{truck}^2 + \frac{1}{2} m_{car} v_{car}^2$

Substituting the values:

$KE_{initial} = \frac{1}{2} (1350)(33,33^2) + \frac{1}{2} (1050)(16,67^2) = 895 741,68 ext{ J}$

Final kinetic energy (KE_final):

$KE_{final} = \frac{1}{2} m_{truck} v_{truck ext{ new}}^2 + \frac{1}{2} m_{car} v_{car ext{ new}}^2$

Calculating with the new velocities:

$KE_{final} = \frac{1}{2} (1350)(20,3^2) + \frac{1}{2} (1050)(-5,32^2) = 293 019,51 ext{ J}$

Since KE_initial ≠ KE_final, the collision is inelastic.

The relationship is inversely proportional. As the contact time increases, the net force experienced by the car decreases, and vice versa.

Impulse is equal to the change in momentum experienced by the car during the crash.

Airbags increase the contact time during a collision. By extending the time over which the force is applied, the average force exerted on the driver decreases, thereby reducing the risk of injury.

Using the formula for impulse:

$F_{net} \cdot \Delta t = \Delta p$

Where $F_{net} = -57 500 ext{ N}, \Delta p = m(v_f - v_i)$

Given that the final velocity $v_f = 0 ext{ m.s}^{-1}$ and initial velocity $v_i = 15 ext{ m.s}^{-1}$, we find:

$\Delta p = 1 150 (0 - 15) = -17 250 ext{ kg.m.s}^{-1}$

Setting the equations equal to one another:

$-57 500 ext{ N} \cdot \Delta t = -17 250$

Thus,

$\Delta t = \frac{-17 250}{-57 500} = 0,30 ext{ s}.$

The contact time during the crash is 0,30 s.